• Lab 1 due date extended 1 week (too many typos!). It is now due 9 October.
• Added 30 minutes to office hours 5-5:30 wed, mostly for this class.
• Will do some more on lab1 in about 10 minutes.

Jesse Wei-Yeh Chu  writes:

> The last program text in Module no. 4 of input set 2 has only 4 digits
> (4999). Is this a typo or an intended error for our linker to catch?

Typo!

> Also for input no. 2, there is a discrepancy between the pdf file and the
> text file. Module 5 in pdf indicates both definition and use list as 00,
> whereas the text version contains only one 00.

The text version is wrong.  A filter I used to save space squashed
duplicate lines (thinking they were blanks).  Alas it doesn't work
here.

I fixed both errors.

End of Note.

What follows is Peterson's solution. When it was published, it was a surprise to see such a simple soluntion. In fact Peterson gave a solution for any number of processes. A proof that the algorithm satisfies our properties (including a strong fairness condition) can be found in Operating Systems Review Jan 1990, pp. 18-22.

Initially P1wants=P2wants=false  and  turn=1

Code for P1                        Code for P2

Loop forever {                     Loop forever {
    P1wants <-- true                   P2wants <-- true
    turn <-- 2                         turn <-- 1
    while (P2wants and turn=2) {}      while (P1wants and turn=1) {}
    critical-section                   critical-section
    P1wants <-- false                  P2wants <-- false
    non-critical-section               non-critical-section

Hardware assist (test and set)

Now implementing a critical section for any number of processes is trivial.

loop forever {
    while (TAS(s)) {}   ENTRY
    CS
    s<--false           EXIT
    NCS

P and V and Semaphores

Note: Tanenbaum does both busy waiting (like above) and blocking (process switching) solutions. We will only do busy waiting.

Homework: 3

The entry code is often called P and the exit code V (Tanenbaum only uses P and V for blocking, but we use it for busy waiting). So the critical section problem is to write P and V so that

loop forever
    P
    critical-section
    V
    non-critical-section

1. Mutual exclusion
2. No speed assumptions
3. No blocking by processes in NCS
4. Forward progress (my weakened version of Tanenbaum's last condition

Note that I use indenting carefully and hence do not need (and sometimes omit) the braces {}

A binary semaphore abstracts the TAS solution we gave for the critical section problem.

• A binary semaphore S takes on two possible values ``open'' and ``closed''
• Two operations are supported
• P(S) is

      while (S=closed) {}
      S<--closed     <== This is NOT the body of the while
  

  where finding S=open and setting S<--closed is atomic
• That is, wait until the gate is open, then run through and atomically close the gate
• Said another way, it is not possible for two processes doing P(S) simultaneously to both see S=open (unless a V(S) is also simultaneous with both of them).
• V(S) is simply S<--open

The above code is not real, i.e., it is not an implementation of P. It is, instead, a definition of the effect P is to have.

To repeat: for any number of processes, the critical section problem can be solved by

loop forever
    P(S)
    CS
    V(S)
    NCS

The only specific solution we have seen for an arbitrary number of processes is the one just above with P(S) implemented via test and set.

Remark: Peterson's solution requires each process to know its processor number. The TAS soluton does not. Thus, strictly speaking Peterson did not provide an implementation of P and V. He did solve the critical section problem.

To solve other coordination problems we want to extend binary semaphores.

• With binary semaphores, two consecutive Vs do not permit two subsequent Ps to succeed (the gate cannot be doubly opened).
• We might want to limit the number of processes in the section to 3 or 4, not always just 1.

The solution to both of these shortcomings is to remove the restriction to a binary variable and define a generalized or counting semaphore.

• A counting semaphore S takes on non-negative integer values
• Two operations are supported
• P(S) is

      while (S=0)
      S--
  

  where finding S>0 and decrementing S is atomic
• That is, wait until the gate is open (positive), then run through and atomically close the gate one unit
• Said another way, it is not possible for two processes doing P(S) simultaneously to both see the same positive value of S unless a V(S) is also simultaneous.
• V(S) is simply S++

These counting semaphores can solve what I call the semi-critical-section problem, where you premit up to k processes in the section. When k=1 we have the original critical-section problem.

initially S=k

loop forever
    P(S)
    SCS   <== semi-critical-section
    V(S)
    NCS

Producer-consumer problem

• Two classes of processes
  • Producers, which produce times and insert them into a buffer.
  • Consumers, which remove items and consume them.
• What if the producer encounters a full buffer?
  Answer: Block it.
• What if the consumer encounters an empty buffer?
  Answer: Block it.
• Also called the bounded buffer problem.
  • Another example of active entities being replaced by a data structure when viewed at a lower level (Finkel's level principle).

Initially e=k, f=0 (counting semaphore); b=open (binary semaphore)

Producer                         Consumer

loop forever                     loop forever
    produce-item                     P(f)
    P(e)                             P(b); take item from buf; V(b)
    P(b); add item to buf; V(b)      V(e)
    V(f)                             consume-item

• k is the size of the buffer
• e represents the number of empty buffer slots
• f represents the number of full buffer slots
• We assume the buffer itself is only serially accessible. That is, only one operation at a time.
  • This explains the P(b) V(b) around buffer operations
  • I use ; and put three statements on one line to suggest that a buffer insertion or removal is viewed as one atomic operation.
  • Of course this writing style is only a convention, the enforcement of atomicity is done by the P/V.
• The P(e), V(f) motif is used to force bounded alternation. If k=1 it gives strict alternation.

Dining Philosophers

A classical problem from Dijkstra

• 5 philosophers sitting at a round table
• Each has a plate of spaghetti
• There is a fork between each two
• Need two forks to eat

• The obvious solution (pick up right; pick up left) deadlocks.
• Big lock around everything serializes.
• Good code in the book.

The purpose of mentioning the Dining Philosophers problem without giving the solution is to give a feel of what coordination problems are like. The book gives others as well. We are skipping these (again this material would be covered in a sequel course). If you are interested look, for example, at http://allan.ultra.nyu.edu/~gottlieb/courses/1997-98-spring/os/class-notes.html.

Homework: 14,15 (these have short answers but are not easy).

Readers and writers

• Two classes of processes.
  • Readers, which can work concurrently.
  • Writers, which need exclusive access.
• Must prevent 2 writers from being concurrent.
• Must prevent a reader and a writer from being concurrent.
• Must permit readers to be concurrent when no writer is active.
• Perhaps want fairness (i.e., freedom from starvation).
• Variants
  1. Writer-priority readers/writers.
  2. Reader-priority readers/writers.

Quite useful in multiprocessor operating systems. The ``easy way out'' is to treat all processes as writers in which case the problem reduces to mutual exclusion (P and V). The disadvantage of the easy way out is that you give up reader concurrency. Again for more information see the web page referenced above.